Help me plz i don’t know the charges on each of these

When illuminated with monochromatic light, a double slit produces a pattern that is a combination of single-slit diffraction and

d1i1m1o1n [39]

Answer:

The ratio is $k:d = 1 : 5$

Explanation:

From the question we are told that

The first minimum of the single slit pattern falls on the fifth maximum of the double slit pattern.

Generally the condition for constructive interference for as single slit is

$ksin(\theta) = n\lambda$

Here k is the width of the slit and n is the order of the fringe and for single slit n = 1 (cause we are considering the first maxima)

Generally the condition for constructive interference for as double slit is

$dsin\theta = m\lambda$

Here d is the separation between the slit and m is the order of the fringe and for double slit m = 5 (cause we are considering the first maxima)

=> $dsin\theta = 5\lambda$

So

$\frac{ksin(\theta)}{dsin(\theta)} = \frac{\lambda}{5\lambda}$

=> $\frac{k }{d} = \frac{1}{5}$

So

$k:d = 1 : 5$

5 0

3 years ago

What area must the plates of a capacitor be if they have a charge of 5.7uC and an electric field of 3.1 kV/mm between them? O 0.

kirill [66]

Answer:

Area of the plates of a capacitor, A = 0.208 m²

Explanation:

It is given that,

Charge on the parallel plate capacitor, $q = 5.7\ \mu C=5.7\times 10^{-6}\ C$

Electric field, E = 3.1 kV/mm = 3100000 V/m

The electric field of a parallel plates capacitor is given by :

$E=\dfrac{q}{A\epsilon_o}$

$A=\dfrac{q}{E\epsilon_o}$

$A=\dfrac{5.7\times 10^{-6}\ C}{3100000\ V/m\times 8.85\times 10^{-12}\ F/m}$

A = 0.208 m²

So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.

8 0

3 years ago

How is sound produced

Dafna11 [192]

Sounds are created when something vibrates (shakes back and forth), sending waves of vibrations into the ears of the listener. When a bell is struck, the metal vibrates. The vibrations travel through the air as sound waves.

hope this helps

7 0

2 years ago

If you neglect air resistance, the horizontal component of a projectile A. Is always changing B. Remains constant throughout the

Shtirlitz [24]

Answer:

B. Remains constant throughout the flight

Explanation:

If we completely neglect air resistance then the projectile will not have any horizontal resistances to deal with, therefore the horizontal component will remain constant throughout the flight. This would continue to be the case until it meets some form of resistance. Which assuming that everything else is normal would be the case since the force of gravity will push the projectile down (vertical component) until it hits the ground which in that instance would act as an opposing force to the horizontal component as well.

8 0

3 years ago

Consider your moment of inertia about a vertical axis through the center of your body, both when you are standing straight up wi

jeka94

Answer:

I₁ / I₂ = 1.43

Explanation:

To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended

Before starting let's reduce all units to the SI system

d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m

d₂ = 38 in = 96.52 10⁻² m

The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm

I₁ = I_man + 2 I_ arm

Man indicates that we can approximate them to a cylinder where the average diameter is

d = (d₁ + d₂) / 2

d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m

The average radius is

r = d / 2 = 50.8 10⁻² m = 0.508 m

The mass of the trunk is the mass of man minus the masses of each arm.

M = M_man - 0.2 M_man = 80 (1-0.2)

M = 64 kg

The moments of inertia are:

A cylinder with respect to a vertical axis: Ic = ½ M r²

A rod that rotates at the end: I_arm = 1/3 m L²

Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.

I1 = I_arm + m D²

Where D is the distance from the axis of rotation of the arm to the axis of the body

D = d / 2 = 101.6 10⁻² /2 = 0.508 m

Let's replace

I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]

Let's calculate

I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]

I₁ = 8.258 + 5.33 + 4.129

I₁ = 17,717 Kg m² / s²

Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,

I₂ = ½ M r² + 2 m D²

I₂ = ½ 64 0.508² + 2 8 0.508²

I₂ = 8,258 + 4,129

I₂ = 12,387 kg m² / s²

The relationship between these two magnitudes is

I₁ / I₂ = 17,717 /12,387

I₁ / I₂ = 1.43

3 0

3 years ago